第37页 - 通城学典课时作业本九年级数学人教版南通专版

C

7

$\sqrt{3}$

$证明:(1)∵ AC^{2}=CE×CB,∴\frac{AC}{CE}=\frac{CB}{AC}$

$又∵ ∠ACB=∠ECA=90°$

$∴△ACB∽△ECA,∴∠ABC=∠EAC$

$∵ D是斜边AB 的中点,∴ CD=AD,∴ ∠ACD=∠CAD$

$∵∠CAD+∠ABC= 90°, ∴ ∠ACD+ ∠EAC = 90°$

$∴∠AFC=90°,∴AE⊥CD$

$(2)∵AE⊥CD,∴∠EFC=90°,∴∠EFC=∠ACE$

$又∵∠CEF=∠AEC,∴△ECF∽△EAC,∴\frac{EC}{EA}=\frac{EF}{EC}$

$∵ E是边BC的中点，∴ CE=BE,∴\frac{BE}{EA}=\frac{EF}{BE}$

$∵ ∠BEF=∠AEB，∴ △BEF∽△AEB,∴∠EBF=∠EAB$

$证明:(1)∵AC平分∠DAB，∴∠DAC=∠CAB$

$∵∠ADC=∠ACB=90°,∴△ADC∽△ACB,∴\frac{AD}{AC}=\frac{AC}{AB},∴ AC^{2}=AB×AD$

$(2)∵∠ACB=90°,E为AB 的中点，∴CE=\frac{1}{2}AB=AE,∴ ∠EAC = ∠ECA$

$∵∠DAC= ∠CAB，∴∠DAC=∠ECA,∴ CE//AD$

$(3)∵ CE//AD,∴△AFD∽△CFE,∴\frac{AD}{CE}=\frac{AF}{CF}$

$∵CE=\frac{1}{2}AB,AB=6,∴CE=3$

$∵ AD=4,∴ \frac{AF}{CF}=\frac{4}{3},∴\frac{CF}{AF}=\frac{3}{4}$

$∴\frac{AC}{AF}=\frac{AF+CF}{AF}=1+\frac{CF}{AF}=\frac{7}{4}$